TS EAMCET 7 May 2018 Shift 1 Solved Paper

The figure below represents the one vertex of square loop at origin,
For wire AB , magnetic filed, B0(1+0)=B0
x=0
For wire CD, magnetic filed, B0(1+

y

L

)
x=L
For wire BC, magnetic filed, B0(1+

x

L

)
y=L
For wire AD, magnetic filed, B0
y=0
The forces in the square loop,
F1=

L

∫

0

IB0=IB0L
F2=

L

∫

0

IB0(1+

x

L

)dx=

3

2

IB0L
F3=

L

∫

0

IB0(1+

x

L

)dx=

3

2

IB0L
F4=

L

∫

0

IB0dx=IB0L
The force in x direction,
Fx=F3−F1
Fx=

3

2

IB0L−IB0L
Fx=

1

2

IB0L
Similarly, The force in y direction,
Fy=F2−F4
Fy=

3

2

IB0L−IB0L
Fy=

1

2

IB0L
The net magnitude of force,
F=√(

1

2

IB0L)2+(

1

2

IB0L)2
F=

IB0L

√2